The sides of a triangle are $x^2 + x + 1$, $2 x + 1$ and $x^2 - 1$. Show that the greatest angle is $120^{\circ}$.
Let the sides of the triangle be
$a = x^2 + x + 1$, $\;$ $b = 2 x + 1$, $\;$ $c = x^2 - 1$
Then, greatest side is side `$a$' and greatest angle is `$A$'.
By cosine rule, $\;$ $\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$
$\begin{aligned}
i.e. \; \cos A & = \dfrac{\left(2 x + 1\right)^2 + \left(x^2 - 1\right)^2 - \left(x^2 + x + 1\right)^2}{2 \left(2 x + 1\right) \left(x^2 - 1\right)} \\\\
& = \dfrac{4 x^2 + 4 x + 1 + x^4 - 2 x^2 + 1 - \left(x^4 + x^2 + 1 + 2 x^3 + 2 x + 2x^2\right)}{2 \left(2x + 1\right) \left(x^2 - 1\right)} \\\\
& = \dfrac{x^4 + 2 x^2 + 4x + 2 - x^4 - 2 x^3 - 3 x^2 - 2 x - 1}{2 \left(2 x^3 - 2 x + x^2 - 1\right)} \\\\
& = \dfrac{- 2 x^3 - x^2 + 2 x + 1}{2 \left(2 x^3 + x^2 - 2 x - 1\right)} \\\\
& = - \dfrac{1}{2}
\end{aligned}$
$\implies$ $A = \cos^{-1} \left(-\dfrac{1}{2}\right) = 120^{\circ}$